Question

A closed combustion chamber is designed so that it maintains a constant pressure of 300 kPa during a combustion process. The combustion chamber has an initial volume of \(0.5 \mathrm{m}^{3}\) and contains a stoichiometric mixture of octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) gas and air at \(25^{\circ} \mathrm{C}\). The mixture is now ignited, and the product gases are observed to be at \(1000 \mathrm{K}\) at the end of the combustion process. Assuming complete combustion, and treating both the reactants and the products as ideal gases, determine the heat transfer from the combustion chamber during this process.

Short answer

Answer: The heat transfer from the combustion chamber during this process is -3.15 x 10^8 J.

Step-by-step solution

Identify known information

First, let's list down the given information: 1. Constant pressure: \(P = 300 \, \mathrm{kPa}\) 2. Initial volume: \(V_1 = 0.5 \, \mathrm{m^3}\) 3. The stoichiometric mixture contains octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) and air 4. The initial temperature: \(T_1 = 25^{\circ}\mathrm{C} = 298.15 \, \mathrm{K}\) 5. The final temperature: \(T_2 = 1000 \, \mathrm{K}\)

Find the balanced chemical equation for the complete combustion of octane

The balanced chemical equation for the combustion of octane is as follows: $$\mathrm{C}_8 \mathrm{H}_{18} + 12.5\left(\mathrm{O}_2 + 3.76\mathrm{N}_2\right) \rightarrow 8 \mathrm{CO}_2 + 9 \mathrm{H}_2\mathrm{O} + 47 \mathrm{N}_2$$

Apply the ideal gas law to determine moles of reactants

We can use the ideal gas law equation, \(PV = nRT\), to find the number of moles (n) for the initial state: $$n_1 = \frac{P_1V_1}{RT_1} = \frac{300,000 \, \mathrm{Pa} \times 0.5 \, \mathrm{m^3}}{8.314 \, \mathrm{\frac{J}{mol\cdot K}}\times 298.15 \, \mathrm{K}} = 60.76 \, \mathrm{mol}$$

Determine the moles of products after combustion

From the balanced chemical equation, we know that complete combustion consumes 12.5 moles of air for each mole of octane. We can write the number of moles of the octane and air as: $$n_{\mathrm{C_8H_{18}}} = \frac{1}{13.5}n_1$$ $$n_{\mathrm{Air}} = \frac{12.5}{13.5}n_1$$ So, we can calculate the moles of products: \(8\mathrm{CO}_2\), \(9\mathrm{H}_2\mathrm{O}\), and \(47\mathrm{N}_2\): $$n_{\mathrm{CO_2}} = 8n_{\mathrm{C_8H_{18}}} = \frac{8}{13.5}n_1$$ $$n_{\mathrm{H_2 O}} = 9n_{\mathrm{C_8H_{18}}} = \frac{9}{13.5}n_1$$ $$n_{\mathrm{N_2}} = 47n_{\mathrm{C_8H_{18}}} = \frac{47}{13.5}n_1$$ Summing up the moles product gives us total product mole \(n_2\): $$n_2 = n_{\mathrm{CO_2}} + n_{\mathrm{H_2 O}} + n_{\mathrm{N_2}} = \frac{64}{13.5}n_1 = 4.74n_1$$

Calculate the final volume using the ideal gas law

Now, we can use the ideal gas law again for the final state: $$V_2 = \frac{n_2RT_2}{P} = \frac{4.74n_1 \times 8.314 \, \mathrm{\frac{J}{mol\cdot K}}\times 1000\, \mathrm{K}}{300,000\, \mathrm{Pa}} = 2.381 \, \mathrm{m^3}$$

Apply energy balance equation to determine heat transfer

We can apply the energy balance equation for the combustion process: $$Q = n_1\Delta H_{\mathrm{rxn}} + \sum n_i\left(C_{v,i}^{\mathrm{product}}\Delta T_i\right)$$ We already know the moles and temperature differences for each product. The heat of the reaction (\(\Delta H_{\mathrm{rxn}}\)) for complete combustion of \(\mathrm{C_8H_{18}}\) is -\(5.47 \times 10^6 \, \mathrm{\frac{J}{mol}}\). Using the specific heat capacities (\(C_{v,i}\)) for each product (\(\mathrm{CO_2}: 821 \, \mathrm{\frac{J}{mol\cdot K}}\), \(\mathrm{H_2O}: 1430\, \mathrm{\frac{J}{mol\cdot K}}\), \(\mathrm{N_2}: 744\, \mathrm{\frac{J}{mol\cdot K}}\)), we can calculate the heat transfer Q: $$Q = 60.76 \times (-5.47 \times 10^6) + \left[\frac{8}{13.5}n_1 \times 821 \times (1000 - 298.15) + \frac{9}{13.5}n_1 \times 1430 \times (1000 - 298.15) + \frac{47}{13.5}n_1 \times 744 \times (1000 - 298.15)\right] = -3.15 \times 10^8 \, \mathrm{J}$$ So, the heat transfer from the combustion chamber during this process is \(-3.15 \times 10^8 \, \mathrm{J}\). The negative sign indicates that the heat is being transferred out of the combustion chamber.